Suppose that, on average, 4 customers arrive each minute in a Poisson distribution. What is the probability that 2 customers will arrive in a particular minute?
A) 0.500
B) 0.147
C) 0.018
D) 0.073
E) 0.195

Question

Quizplus · Accepted Answer

The probability of observing $k$ events in a fixed interval in a Poisson distribution is given by $P(k) = \frac{\lambda^k e^{-\lambda}}{k!}$, where $\lambda$ is the average rate of occurrence. Here, $\lambda = 4$ customers per minute, and we want the probability of $k = 2$ customers arriving. Thus, $P(2) = \frac{4^2 e^{-4}}{2!} = \frac{16 e^{-4}}{2} = 0.1465$, which rounds to 0.147.

Suppose That, on Average, 4 Customers Arrive Each Minute in a Poisson